PROOF If His normal in G, then for any x[Gand h[Hthere is an h9 in Hsuch that xh5h9x. Thus,xhx215h9, and therefore xHx21# H.
Conversely, if xHx21 # H for all x, then, letting x 5 a, we have aHa21# H or aH# Ha. On the other hand, letting x5a21, we have a21H(a21)215a21Ha # H or Ha # aH.
EXAMPLE 1 Every subgroup of an Abelian group is normal. (In this case,ah5hafor ain the group and hin the subgroup.)
EXAMPLE 2 The center Z(G) of a group is always normal. [Again, ah5hafor any a[Gand any h[Z(G).]
EXAMPLE 3 The alternating group Anof even permutations is a nor- mal subgroup of Sn. [Note, for example, that for (12) [Snand (123) [ An, we have (12)(123) 2 (123)(12) but (12)(123) 5 (132)(12) and (132) [An.]
EXAMPLE 4 The subgroup of rotations in Dnis normal in Dn. (For any rotation rand any reflection f, we have fr5r21f, whereas for any rotations rand r9, we have rr9 5r9r.)
EXAMPLE 5 The group SL(2,R) of 2 32 matrices with determinant 1 is a normal subgroup of GL(2,R), the group of 2 32 matrices with nonzero determinant. To verify this, we use the normal subgroup test given in Theorem 9.1. Let x[GL(2,R) 5G,h[SL(2,R) 5Hand note that det xhx215(det x)(det h)(det x)215(det x)(det x)2151. So, xhx21[H, and, therefore,xHx21#H.
EXAMPLE 6 Referring to the group table for A4given in Table 5.1 on page 107, we may observe that H 5 {a1, a2, a3, a4} is a normal subgroup of A4, whereas K 5 {a1, a5, a9} is not a normal subgroup ofA4. To see that His normal, simply note that for any bin A4,bHb21is a subgroup of order 4 and H is the only subgroup of A4 of order 4 since all other elements of A4have order 3. Thus,bHb215H. In con- trast,a2a5a212 5 a7, so that a2Ka221sK.
A subgroup H of G is normal in G if and only if xHx21#H for all x in G.
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Factor Groups
We have yet to explain why normal subgroups are of special significance.
The reason is simple. When the subgroup Hof Gis normal, then the set of left (or right) cosets of Hin Gis itself a group—called the factor group of G by H(or the quotient group of G by H). Quite often, one can obtain information about a group by studying one of its factor groups. This method will be illustrated in the next section of this chapter.
Theorem 9.2 Factor Groups (O. Hölder, 1889)
PROOF Our first task is to show that the operation is well defined; that is, we must show that the correspondence defined above from G/H3 G/Hinto G/H is actually a function. To do this we assume that for some elements a,a9,b, and b9from G, we have aH5 a9Hand bH5 b9Hand verify that aHbH5a9Hb9H. That is, verify that abh5a9b9H.
(This shows that the definition of multiplication depends only on the cosets and not on the coset representatives.) From aH5a9Hand bH5 b9H, we have a9 5ah1and b9 5bh2for some h1,h2in H, and therefore a9b9H5ah1bh2H5ah1bH5ah1Hb5aHb5abH. Here we have made multiple use of associativity, property 2 of the lemma in Chapter 7, and the fact that HvG. The rest is easy:eH5His the identity; a21His the inverse of aH; and (aHbH)cH 5 (ab)HcH 5 (ab)cH 5 a(bc)H 5 aH(bc)H5aH(bHcH). This proves that G/His a group.
Although it is merely a curiosity, we point out that the converse of Theorem 9.2 is also true; that is, if the correspondence aHbH5abH defines a group operation on the set of left cosets of Hin G, then His normal in G.
The next few examples illustrate the factor group concept.
EXAMPLE 7 Let 4Z5{0,64,68, . . .}. To construct Z/4Z, we first must determine the left cosets of 4Zin Z. Consider the following four cosets:
0 14Z54Z5{0,64,68, . . .},
1 14Z5{1, 5, 9, . . . ; 23,27,211, . . .}, Let G be a group and let H be a normal subgroup of G. The set G/H 5{aH |a [G} is a group under the operation (aH)(bH)5abH.†
†The notation G/Hwas first used by C. Jordan.
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9 | Normal Subgroups and Factor Groups 181
2 14Z5{2, 6, 10, . . . ; 22,26,210, . . .}, 3 14Z5{3, 7, 11, . . . ; 21,25,29, . . .}.
We claim that there are no others. For if k[Z, then k54q1r, where 0 #r,4; and, therefore,k14Z5r14q14Z5r14Z. Now that we know the elements of the factor group, our next job is to determine the structure of Z/4Z. Its Cayley table is
0 14Z 1 14Z 2 14Z 3 14Z
014Z 0 14Z 1 14Z 2 14Z 3 14Z
114Z 1 14Z 2 14Z 3 14Z 0 14Z
214Z 2 14Z 3 14Z 0 14Z 1 14Z
314Z 3 14Z 0 14Z 1 14Z 2 14Z
Clearly, then,Z/4ZLZ4. More generally, if for any n.0 we let nZ5 {0,6n,62n,63n, . . .}, then Z/nZis isomorphic to Zn.
EXAMPLE 8 Let G5Z18and let H565{0, 6, 12}. Then G/H5 {0 1H, 1 1H, 2 1H, 3 1H, 4 1H, 5 1H}. To illustrate how the group elements are combined, consider (5 1 H) 1 (4 1 H). This should be one of the six elements listed in the set G/H. Well, (5 1H) 1 (4 1H) 55 14 1H59 1H53 16 1H53 1H, since Hab- sorbs all multiples of 6.
A few words of caution about notation are warranted here. When H is a normal subgroup of G, the expression |aH|has two possible inter- pretations. One could be thinking of aHas a setof elements and |aH|
as the size of the set; or, as is more often the case, one could be think- ing of aHas a group element of the factor group G/Hand |aH|as the order of the element aHin G/H. In Example 8, for instance, the set3 1 H has size 3, since 3 1 H 5 {3, 9, 15}. But the group element 3 1Hhas order 2, since (3 1H) 1(3 1H) 56 1H50 1H. As is usually the case when one notation has more than one meaning, the ap- propriate interpretation will be clear from the context.
EXAMPLE 9 Let _5{R0, R180}, and consider the factor group of the dihedral group D4(see page 31 for the multiplication table for D4)
D4/_5{_,R90_,H_,D_}.
The multiplication table for D4/_is given in Table 9.1. (Notice that even though R90H 5D9, we have used D_in Table 9.1 for H_R90_ because D9_5D_.)
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D4/_ provides a good opportunity to demonstrate how a factor group of Gis related to Gitself. Suppose we arrange the heading of the Cayley table for D4in such a way that elements from the same coset of _are in adjacent columns (Table 9.2). Then, the multiplication table for D4can be blocked off into boxes that are cosets of _, and the sub- stitution that replaces a box containing the element xwith the coset x_
yields the Cayley table for D4/_.
For example, when we pass from D4to D4/_, the box
in Table 9.2 becomes the element H_in Table 9.1. Similarly, the box
becomes the element D_, and so on.
Table 9.2
R0 R180 R90 R270 H V D D9 R0 R0 R180 R90 R270 H V D D9 R180 R180 R0 R270 R90 V H D9 D R90 R90 R270 R180 R0 D9 D H V R270 R270 R90 R0 R180 D D9 V H H H V D D9 R0 R180 R90 R270 V V H D9 D R180 R0 R270 R90 D D D9 V H R270 R90 R0 R180 D9 D9 D H V R90 R270 R180 R0
H V
V H
D D9 D9 D
Table 9.1 _
_ R90__ H__ D__ _
_ _ R90_ H_ D_
R90__ R90_ _ D_ H_
H__ H_ D_ _ R90_
D__ D_ H_ R90_ _
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9 | Normal Subgroups and Factor Groups 183
Table9.3
1 2 3 4 5 6 7 8 9 10 11 12
1 1 2 3 4 5 6 7 8 9 10 11 12
2 2 1 4 3 6 5 8 7 10 9 12 11
3 3 4 1 2 7 8 5 6 11 12 9 10
4 4 3 2 1 8 7 6 5 12 11 10 9
5 5 8 6 7 9 12 10 11 1 4 2 3
6 6 7 5 8 10 11 9 12 2 3 1 4
7 7 6 8 5 11 10 12 9 3 2 4 1
8 8 5 7 6 12 9 11 10 4 1 3 2
9 9 11 12 10 1 3 4 2 5 7 8 6
10 10 12 11 9 2 4 3 1 6 8 7 5
11 11 9 10 12 3 1 2 4 7 5 6 8
12 12 10 9 11 4 2 1 3 8 6 5 7
Table 9.4
1H 5H 9H
1H 1H 5H 9H
5H 5H 9H 1H
9H 9H 1H 5H
In this way, one can see that the formation of a factor group G/H causes a systematic collapse of the elements of G. In particular, all the elements in the coset of Hcontaining acollapse to the single group el- ement aHin G/H.
EXAMPLE10 Consider the group A4as represented by Table 5.1 on page 107. (Here idenotes the permutation ai.) Let H5{1, 2, 3, 4}.
Then the three cosets of Hare H, 5H5{5, 6, 7, 8}, and 9H5{9, 10, 11, 12}. (In this case, rearrangement of the headings is unneces- sary.) Blocking off the table for A4into boxes that are cosets of H and replacing the boxes containing 1, 5, and 9 (see Table 9.3) with the cosets 1H, 5H, and 9H, we obtain the Cayley table for G/Hgiven in Table 9.4.
This procedure can be illustrated more vividly with colors. Let’s say we had printed the elements of Hin green, the elements of 5Hin red, and the elements of 9Hin blue. Then, in Table 9.3, each box would consist of elements of a uniform color. We could then think of
184 Groups
the factor group as consisting of the three colors that define a group table isomorphic to G/H.
It is instructive to see what happens if we attempt the same proce- dure with a group Gand a subgroup Hthat is not normal in G—that is, if we arrange the headings of the Cayley table so that the elements from the same coset of Hare in adjacent columns and attempt to block off the table into boxes that are also cosets of Hto produce a Cayley table for the set of cosets. Say, for instance, we were to take Gto be A4 and H5 {1, 5, 9}. The cosets of Hwould be H, 2H5 {2, 6, 10}, 3H5{3, 7, 11}, and 4H5{4, 8, 12}. Then the first three rows of the rearranged Cayley table for A4would be
Green Red Blue
Green Green Red Blue
Red Red Blue Green
Blue Blue Green Red
1 5 9 2 6 10 3 7 11 4 8 12
1 1 5 9 2 6 10 3 7 11 4 8 12
5 5 9 1 8 12 4 6 10 2 7 11 3
9 9 1 5 11 3 7 12 4 8 10 2 6
But already we are in trouble, for blocking these off into 3 33 boxes yields boxes that contain elements of different cosets. Hence, it is im- possible to represent an entire box by a single element of the box in the same way we could for boxes made from the cosets of a normal sub- group. Had we printed the rearranged table in four colors with all members of the same coset having the same color, we would see multi- colored boxes rather than the uniformly colored boxes produced by a normal subgroup.
In Chapter 11, we will prove that every finite Abelian group is isomorphic to a direct product of cyclic groups. In particular, an Abelian group of order 8 is isomorphic to one of Z8,Z4%Z2, or Z2% Z2%Z2. In the next two examples, we examine Abelian factor groups of order 8 and determine the isomorphism type of each.
EXAMPLE11 Let G5U(32) 5{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31} and H5U16(32) 5{1, 17}. Then G/His an Abelian group of order 16/2 58. Which of the three Abelian groups of order 8 is it—Z8,Z4%Z2, or Z2%Z2%Z2? To answer this question, we need
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